We prove (REF ) in two steps. First we show that \(\widehat{W}(t)/n\)  is a tight sequence in D[0,\(\widehat{s}(0)-\epsilon \) ], the space of RCLL functions on the interval [0,\(\widehat{s}(0)-\epsilon \) ]. Then we show the uniqueness of possible sequential limit.
Since \(\widehat{W}(0) \rightarrow \widehat{w}(0)\) , we know \(\lbrace  \widehat{W}(0)/n \rbrace _{n\ge 1}\)  is a tight sequence of random variables. To establish tightness of \(\lbrace  \widehat{W}(t)/n ,0\le t\le 1-\epsilon \rbrace _{n\ge 1}\) , we need to show that
for any fixed \(\epsilon ^{\prime },\delta >0\) , there exist \(\theta >0\)  and an integer \(n_0\)  so that for all \(n\ge n_0\) ,
\(\mathbb {P}\left(\sup _{\left|t_1-t_2\right|\le \theta , t_1<t_2\le \widehat{s}(0)-\epsilon } \left| \frac{\widehat{W}(t_1)}{n} - \frac{\widehat{W} (t_2)}{n} \right| \ge \delta \right)\le \epsilon ^{\prime }.\) 
Assuming (REF ) for the moment,
we see that \(\lbrace \widehat{W}(t)/n,0\le t\le \widehat{s}(0)-\epsilon \rbrace \) , as an element of D[0,\(\widehat{s}(0)-\epsilon \) ], satisfies condition (ii) of
Proposition 3.26 in [1]}. Consequently,
\(\lbrace \widehat{W}(t)/n,0\le t\le \widehat{s}(0)-\epsilon \rbrace _{n\ge 1}\)  is a tight sequence.
Now we show that any sequential limit of \(\widehat{W}(t)/n\)  coincides with \(\widehat{w}(t)\) .
By the tightness of \(\lbrace \widehat{W}(t)/n,t\ge 0\rbrace _{n\ge 1}\) , we see for any subsequence of \(\widehat{W}(t)/n\)  we can extract a further subsequence that converges in distribution to a process \(\check{w}(t)\)  with continuous sample path. By the Skorokhod representation theorem we can assume the convergence is actually in the almost sure sense, and we can also assume that
\(\widehat{S}(t)/n\)  converges a.s. to \(\widehat{s}(t)\)  in the interval [0,\(\widehat{s}(0)-\epsilon \) ] and \(M_2(t)/n\)  converges to 0 a.s..
It remains to prove that the limit point \(\check{w}(t)\)  coincides with \(\widehat{w}(t)\) , which is then necessarily independent of the subsequence. But this is clearly the case. Indeed, by dividing both sides of (REF ) by \(n\)  and then sending \(n\)  to \(\infty \) , we obtain that
\(\check{w}(t)=\widehat{w}(0)+\int _0^t \frac{\omega }{\lambda }\alpha \widehat{s}(u)du-\int _0^t 2\frac{\check{w}(u)}{\widehat{s}(u)}du,\) 
for \(t\le \widehat{s}(0)-\epsilon \) . This equation has a unique solution \(\check{w}(t)=\widehat{w}(t)\) .
It remains to prove (REF ).
We have
\(\frac{\widehat{W}(t_2)}{n}-\frac{\widehat{W}(t_1)}{n}=\int _{t_1}^{t_2}\left(\frac{\alpha \omega \widehat{S}(u) }{\lambda n} -\frac{2\widehat{W}(u)}{\widehat{S}(u)}\right)du +\frac{M_2(t_2)-M_2(t_1)}{n}.\) 
We define the event
\(\Omega _{\ref {s>ep}}:=\lbrace \widehat{S}^{n}(t)\ge \epsilon n/2 ,\forall t\le (\widehat{s}(0)-\epsilon )\wedge \widehat{\tau } \rbrace  \cap \lbrace  \sup _{t \le \widehat{\tau } \wedge 2}\left|M_2(t)\right|\le n^{2/3} \rbrace .\) 
By (REF ) and (REF ), \(\mathbb {P}(\Omega _{\ref {s>ep}}) \rightarrow 1\)  as \(n\rightarrow \infty \) .
On \(\Omega _{\ref {s>ep}}\) , for \(t_1<t_2<\widehat{s}(0)-\epsilon \) , using (REF ),
\(\begin{split}&\left|\frac{\widehat{W}(t_2)}{n}-\frac{\widehat{W}(t_1)}{n}\right|\\\le &(t_2-t_1)\left(\frac{\alpha \omega }{\lambda }+\frac{2\mu n}{\epsilon n/2}\right)+\frac{2}{n} \sup _{t \le \widehat{\tau } \wedge 1}\left|M_2(t)\right|\\\le & C\left((t_2-t_1)+n^{-1/3}\right).\end{split}\) 
This proves (REF ) and thus also completes the proof of (REF ). .
